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JEE MATHEMATICS MCQ: DEFINITE INTEGRATION

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Application +4 | -1
Let f:R→Rf: R \rightarrow R defined as f(x)=x+sin⁡xf(x)=x+\sin x then value of ∫02πf−1(x)dx=\int\limits_{0}^{2 \pi} f^{-1}(x) d x=
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Correct option: A
2π22 \pi^2
Option D ((x-4)²+(y-5)²=25) is incorrect. Option D is incorrect because the incentre is (1,1) which does not lie on the circle (x-4)²+(y-5)²=25. The incentre of the triangle formed by the lines xy=0 and x/3+y/4=1 is (1,1).

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